Let a graph $G$ be arbitrarily traversable from a vertex $v$, i.e., any trail in $G$ initiatng from $v$ ultimately results in an Eulerian $v-v$ circuit. Let $v$ be a cut-vertex in $G$. Is it true that $v$ is necessarily the only cut-vertex in $G$?
When $v$ is a cut vertex, it means that $G-v$ must be disconnected. Now, we should prove that for any other verex $v_0$ in $G$, the graph $G-v_0$ is connected. I think we should take two vertices $u$ and $w$ in $G-v_0$ and using a certain trail linking these two with $v$ reach at a path between $u$ and $w$. Am I rightly starting? Thanks for any help!